Why is the r-value independent of translation and scaling? #2

If you missed part 1, you should read that first. Lets look at why r-value is independent of translation and scaling from the mathematical approach.

Credits: MF15

Credits: MF15

Consider x=aw+b so a is a scaling parameter while b is a translation parameter. Notice first we can re-write the formula like this.

\frac {\sum(x-\sum(\frac{x}{n}))(y-\bar{y})}{\sqrt{[\sum(x-\sum{\frac{x}{n}})^2][\sum(y-\bar{y})^2]}}&s=3

=\frac {\sum({aw+b}-\sum(\frac{aw+b}{n}))(y-\bar{y})}{\sqrt{[\sum({aw+b}-\sum{\frac{aw+b}{n}})^2][\sum(y-\bar{y})^2]}}&s=3

=\frac {\sum({aw+b}-\sum(\frac{aw}{n})-b)(y-\bar{y})}{\sqrt{[\sum({aw+b}-\sum{\frac{aw}{n}}-b)^2][\sum(y-\bar{y})^2]}}&s=3

=\frac {\sum({aw}-\sum(\frac{aw}{n}))(y-\bar{y})}{\sqrt{[\sum({aw}-\sum{\frac{aw}{n}})^2][\sum(y-\bar{y})^2]}}&s=3

=\frac {\sum({aw}-a\sum(\frac{w}{n}))(y-\bar{y})}{\sqrt{[\sum({aw}-a\sum{\frac{w}{n}})^2][\sum(y-\bar{y})^2]}}

=\frac {\sum({aw}-a\bar{w})(y-\bar{y})}{\sqrt{[\sum({aw}-a\bar{w})^2][\sum(y-\bar{y})^2]}}

=\frac {a\sum(w-\bar{w})(y-\bar{y})}{\sqrt{[a^{2}\sum(w-\bar{w})^2][\sum(y-\bar{y})^2]}}

=\frac {a\sum(w-\bar{w})(y-\bar{y})}{\sqrt{a^2} \sqrt{\sum(w-\bar{w})^2][\sum(y-\bar{y})^2]}}

=\frac {a\sum(w-\bar{w})(y-\bar{y})}{a\sqrt{\sum(w-\bar{w})^2][\sum(y-\bar{y})^2]}}

=\frac {\sum(w-\bar{w})(y-\bar{y})}{\sqrt{\sum(w-\bar{w})^2][\sum(y-\bar{y})^2]}}

We finally end up with the conclusion that the r-value will have no change at all! I hope this explains why r-value is independent of scaling and translation. Do leave comments should you have other methods!

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